= Solution
The source potential rises from zero to $U_0$ over the half-turn from $s$ to $s+\pi$, then falls over the next half-turn. Its slopes are $\pm U_0/\pi$. <Torque> is minus the angular <derivative> of potential; in the detached state $U_B$ is angle-independent and exerts no <torque>. A uniform set of motor phases therefore gives
$$
\boxed{T_0=\frac{U_0}\pi,\qquad
\overline T(s)=\frac{T_0}{2\pi}
\left[-\int_s^{s+\pi}P_A(\theta)d\theta+
\int_{s+\pi}^{s+2\pi}P_A(\theta)d\theta\right].}
$$
Here $U_0$ is an <energy> and $T_0$ a <torque> because the angle is in radians. The average over a full turn divides by $2\pi$, rather than by the bound-state occupation alone.
Differentiating the moving integration limits and using <periodicity> gives
$$
\frac{d\overline T}{ds}=\frac{T_0}{\pi}[P_A(s)-P_A(s+\pi)]
=\frac{T_0}{\pi}[2P_A(s)-1].
$$
The <optimal phase of a two-state rotary motor> follows on $0<s<\pi$: $P_A$ decreases, so the stationary point is a maximum and satisfies
$$
\boxed{P_A(s_{\max})=P_A(s_{\max}+\pi)=\tfrac12,\qquad
s_{\max}=\frac1\beta\ln(2\alpha)
=\frac\pi2-\frac1\beta\ln\cosh\frac{\pi\beta}{2}.}
$$
The crossing displaced by $\pi$ is the corresponding minimum. For slow positive rotation, $s_{\max}\simeq(\omega/k_0)\ln2$ lies just after the detachment switch. The potential then rises mainly while the motor is detached and falls mainly while it is bound. For fast rotation, $s_{\max}\simeq\pi/2-\pi^2k_0/(8\omega)$, with only a small phase-dependent occupation available to produce <torque>. The potential should still be shifted by this optimum, rather than aligned arbitrarily with a rate switch.
The half-turn symmetry gives $\int_s^{s+2\pi}P_A\,d\theta=\pi$. At the optimum, direct integration of the two exponential pieces gives $\int_s^{s+\pi}P_A\,d\theta=s_{\max}$. Thus a useful exact check is
$$
\boxed{\overline T_{\max}=\frac{T_0}{\pi\beta}\ln\cosh\frac{\pi\beta}{2}
\simeq\frac{T_0}{2}-\frac{T_0\ln2}{\pi\beta}\longrightarrow\frac{T_0}{2}
\quad(\omega\ll k_0).}
$$
For fast rotation, $\overline T_{\max}\simeq T_0\pi k_0/(8\omega)$ instead tends to zero. The original plots below show the calculated <probabilities> and optimized potential shapes. The imposed switching rates constitute driven kinetics: their strictly one-way windows are not <equilibrium> <detailed balance> at a fixed angle. Sustained work requires the motor's chemical/ion-gradient <energy> supply, not extraction of work from a passive single thermal bath.
\Image[/past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2018-rotor-probability-and-torque.png]
{title=Motor occupation, optimal angular potential phase and the speed dependence of available torque}
{description=Original solutions of the two-state advective motor model. Potential is normalized by $U_0$, torque by $T_0$, and the slow/fast profiles use illustrative values of $k_0/\omega$.}
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